Question

attempt 1: 10 attempts remaining. evaluate $\frac{d}{dt}(3e^{-11t})$. $\frac{d}{dt}(3e^{-11t})=$

Explanation:

Step1: Use constant - multiple rule

The constant - multiple rule of differentiation states that if \(y = cf(x)\), then \(y'=c\cdot f'(x)\), where \(c\) is a constant. Here \(c = 3\) and \(f(t)=e^{-11t}\), so \(\frac{d}{dt}(3e^{-11t})=3\frac{d}{dt}(e^{-11t})\).

Step2: Use chain - rule for \(e^{-11t}\)

The chain - rule states that if \(y = e^{u}\) and \(u = g(t)\), then \(\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}\). For \(y = e^{u}\) with \(u=-11t\), \(\frac{dy}{du}=e^{u}\) and \(\frac{du}{dt}=- 11\). So \(\frac{d}{dt}(e^{-11t})=e^{-11t}\cdot(-11)\).

Step3: Combine results

Since \(\frac{d}{dt}(3e^{-11t})=3\frac{d}{dt}(e^{-11t})\) and \(\frac{d}{dt}(e^{-11t})=-11e^{-11t}\), then \(\frac{d}{dt}(3e^{-11t})=3\times(-11)e^{-11t}=-33e^{-11t}\).

Answer:

\(-33e^{-11t}\)